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Thread: Inequality involving square roots

  1. #1
    Member Ruun's Avatar
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    Inequality involving square roots

    Hi, I've found a proof from $\displaystyle |x| \leq \sum_{i=1}^{n} |x_{i}|$ given $\displaystyle x=(x_{1},...,x_{n}), n \in \mathbb{N}$ It reads:

    $\displaystyle \sum_{i=1}^{n}(x_{i})^{2} \leq \sum_{i=1}^{n} (x_{i})^2 + 2\sum_{i \ne j}^{n} |x_{i}||x_{j}| = (\sum_{i=1}^{n}|x_{i}|)^2$.

    And taking the square root it's supposed to grant the result.

    Now, I do not see why $\displaystyle a^2 \geq b^2 \Rightarrow a \geq b$

    Thanks in advance
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  2. #2
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    The statement $\displaystyle a^2 \ge b^2 \implies a \ge b $ is equivalent to $\displaystyle a < b \implies a^2 < b^2$ since they are contrapositives. Is it clear to you that $\displaystyle a < b \implies a^2 < b^2$?
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  3. #3
    Member Ruun's Avatar
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    Given $\displaystyle a \leq b \Rightarrow a - b \leq 0$ Now I need information for $\displaystyle a+b$ because $\displaystyle (a+b)(a-b) = a^2-b^2$, and that's the piece that I need.

    I'm sorry but I still don't get it. Thanks for your post.
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  4. #4
    A Plied Mathematician
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    Looking at this idea from a calculus perspective: the square root function is everywhere increasing and positive on its domain. Therefore, it preserves order.
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  5. #5
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    Well what you have written is fine. Given $\displaystyle a < b$ we have that $\displaystyle a-b < 0$ now provided $\displaystyle a + b \ge 0$ we have that $\displaystyle (a^2 -b^2) = (a+b)(a-b) < (a+b) 0 = 0.$
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  6. #6
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    Quote Originally Posted by Ruun View Post
    Now, I do not see why $\displaystyle a^2 \geq b^2 \Rightarrow a \geq b$
    I do not understand what exactly you are asking for.
    However, $\displaystyle a^2 \geq b^2 \Rightarrow a \geq b$ is clearly false.
    It is true that $\displaystyle a^2 \geq b^2 \Rightarrow |a| \geq |b|$.

    Is it also the case that $\displaystyle \left| x \right| = \sqrt {\sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } } ~?$
    If so, the proof seems to work.
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  7. #7
    Member Ruun's Avatar
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    Yes, it is the case.

    Well the proof says taking square root, and I didn't find an argument to believe in that, and this is the point I'm looking for,
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  8. #8
    A Plied Mathematician
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    Reply to Plato @ Post #6:

    In this case, the things we're taking the square roots of are all positive. The step in question is this one:

    $\displaystyle \sum_{i=1}^{n}(x_{i})^{2} \leq (\sum_{i=1}^{n}|x_{i}|)^2$

    implies

    $\displaystyle \sqrt{\sum_{i=1}^{n}(x_{i})^{2}} \leq \sum_{i=1}^{n}|x_{i}|.$

    It is also true that $\displaystyle |x|=\sqrt{\sum_{i=1}^{n}(x_{i})^{2}}$, at least in the Euclidean or $\displaystyle L^{2}$ norm.
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